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Using integration, how would I solve f(t) convolve g(t) given that

$$f(t)=u(t)-u(t-5)$$

and

$$g(t)=2[u(t)-u(t-1)]$$

I know it should be

$$\int_0^6 f(\tau) \ast g(t-\tau)~ d\tau = \int_0^6(u(\tau)-u(t-\tau-5))*(2(u(t-\tau)-u(t-\tau-1)))d\tau$$

but I have no clue where to go from there.

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1 Answer 1

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Just expand term-by-term : $$(f*g)(t)=\int_0^6f(x)g(t-x)dx$$ $$=2\int_0^6[u(x)u(t-x)-u(x-5)u(t-x)-u(x)u(t-x-1)+u(x-5)u(t-x-1)]$$ $$=2(u(x)*u(x)-u(x-5)*u(x)-u(x)*u(x-1)+u(x-5)*u(x-1))$$ $$=2(tu(t)-(t-5)u(t-5)-(t-1)u(t-1)+(t-6)u(t-6))$$

Actually we don't have to expand the terms since convolution is linear operation and I used $u(x-a)*u(x-b)=(t-a-b)u(t-a-b)$.

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